mr. bartelt presents how fast??? chapter 12 chemical kinetics

Mr. Bartelt presentsMr. Bartelt presents

How fast???How fast???

Chapter 12Chapter 12

Chemical KineticsChemical Kinetics

Chemical kineticsChemical kinetics

Thus far we have looked exclusively Thus far we have looked exclusively at reactions from a before and after at reactions from a before and after stand point.stand point.

Now we look at the reactions as they Now we look at the reactions as they happen. This should help complete happen. This should help complete the picture.the picture.

Will it happen?Will it happen?

Look at the equation below.Look at the equation below.

2NO2NO2(g)2(g) 2NO 2NO(g)(g) + O + O2(g)2(g)

We can use We can use ΔΔH and H and ΔΔS to determine S to determine if the reaction will occur at room if the reaction will occur at room temperature.temperature.

Let’s do that:Let’s do that:

HHff of NO of NO2(g)2(g) = 34, S = 240 = 34, S = 240

HHff of NO of NO(g)(g) = 90, S = 211 = 90, S = 211

SHIVA SHIVA or or 2NO2NO2(g)2(g) 2NO 2NO(g)(g) + O + O2(g)2(g)

HHff of NO of NO2(g)2(g) = 34, S = 240 = 34, S = 240

HHff of NO of NO(g)(g) = 90, S = 211 = 90, S = 211

S of OS of O22 = 205 = 205ΔΔH = (2*90)-(2*34)H = (2*90)-(2*34)ΔΔH = 112 kJ/mol H = 112 kJ/mol ΔΔS = (2*211+205)-(2*240)S = (2*211+205)-(2*240)ΔΔS = (627)-(480)=147 J/mol•K S = (627)-(480)=147 J/mol•K ΔΔS = 147 J/mol•K = 0.147 kJ/mol•KS = 147 J/mol•K = 0.147 kJ/mol•K

Temperature controls itTemperature controls it


ΔΔH = 112 kJ/mol H = 112 kJ/mol ΔΔS = 0.147 kJ/mol•K S = 0.147 kJ/mol•K ΔΔG = G = ΔΔH - T H - T ΔΔSS

ΔΔG = 112 – (20+273)(0.147)G = 112 – (20+273)(0.147)

ΔΔG = 112 – (43) = 69 kJ/mol G = 112 – (43) = 69 kJ/mol This will need to be heated or forced This will need to be heated or forced by other means to occur.by other means to occur.

StoichiometryStoichiometry


Thermodynamics tells us if it will Thermodynamics tells us if it will happen.happen.Stoichiometry tells us how much of Stoichiometry tells us how much of each product/reactant is each product/reactant is needed/produced.needed/produced.For instance, we know that 4 moles For instance, we know that 4 moles of NOof NO22 will produce 4 moles of NO will produce 4 moles of NO and 2 moles of Oand 2 moles of O22..

What we don’t know…What we don’t know…

We have no idea how long it will take We have no idea how long it will take those 4 moles of NOthose 4 moles of NO22 to react to react completely.completely.

It doesn’t happen instantly, it takes It doesn’t happen instantly, it takes time.time.

Look at the sample data on the Look at the sample data on the following page.following page.


From now on From now on [bracket] will be [bracket] will be used to express used to express concentration (M).concentration (M).

Note: “S” tells us Note: “S” tells us that [NOthat [NO22] + [NO] ] + [NO] = [NO= [NO22]]initialinitial

[O[O22] = [NO]/2] = [NO]/2

Graphs to comeGraphs to come

Time(sec) [NO2] [NO] [O2]

0 0.0100 0.0000 0.0000

50 0.0079 0.0021 0.0011

100 0.0065 0.0035 0.0018

150 0.0055 0.0045 0.0023

200 0.0048 0.0052 0.0026

250 0.0043 0.0057 0.0029

300 0.0038 0.0062 0.0031

350 0.0034 0.0066 0.0033

400 0.0031 0.0069 0.0035

The graphThe graph

Concentration vs. Time

0.0000

0.0020

0.0040

0.0060

0.0080

0.0100

0.0120

0 100 200 300 400 500

Time (sec)

[co

nce

ntr

atio

n]

(mo

l/L

0

[NO2] [NO] [O2]

Back to mathBack to mathThe rate of reaction is defined as the The rate of reaction is defined as the change in concentration vs. the change in concentration vs. the change in time.change in time.Rate of reaction is the slope of the Rate of reaction is the slope of the line on the graph.line on the graph.Note: Slope will always be expressed Note: Slope will always be expressed as a positive number because a as a positive number because a negative rate of reaction doesn’t negative rate of reaction doesn’t have a lot of physical signficancehave a lot of physical signficance

Rate calculationRate calculationRate of oxygen Rate of oxygen production:production:

From 0From 0 50 sec 50 sec

Time(sec) [NO2] [NO] [O2]

0 0.0100 0.0000 0.0000

50 0.0079 0.0021 0.0011

100 0.0065 0.0035 0.0018

150 0.0055 0.0045 0.0023

200 0.0048 0.0052 0.0026

250 0.0043 0.0057 0.0029

300 0.0038 0.0062 0.0031

350 0.0034 0.0066 0.0033

400 0.0031 0.0069 0.0035

510*2.250

0011.0

050

0000.00011.0

This is the average This is the average rate of oxygen rate of oxygen production from time production from time 00 50 sec. 50 sec.

This is also the This is also the “instantanious rate” “instantanious rate” for time t=25 secfor time t=25 sec

Rate vs. TimeRate vs. TimeRate vs. time lies at Rate vs. time lies at the heart of chemical the heart of chemical kinetics.kinetics.We need to know how We need to know how the rate of the the rate of the reaction is affected reaction is affected by time.by time.Note: the rate of Note: the rate of reaction slows as reaction slows as time moves forward.time moves forward.Graph on next slideGraph on next slide

Time (sec) [O2] Time

Inst. Rate

0 0.0000 25 2.10E-05

50 0.0011 75 1.40E-05

100 0.0018 125 1.00E-05

150 0.0023 175 7.00E-06

200 0.0026 225 5.00E-06

250 0.0029 275 5.00E-06

300 0.0031 325 4.00E-06

350 0.0033 375 3.00E-06

400 0.0035

Why the slow downWhy the slow down

Recall that the thermo data predicts that this reaction Recall that the thermo data predicts that this reaction will be spontaneous in the opposite direction.will be spontaneous in the opposite direction.Hence we force the reaction forward by chemical means. Hence we force the reaction forward by chemical means. But once NO and O But once NO and O22 are produced the will react to form are produced the will react to form NONO22 in the reverse reaction. in the reverse reaction.

Inst. Rate vs. Time

0.00E+00

5.00E-06

1.00E-05

1.50E-05

2.00E-05

2.50E-05

0 100 200 300 400

Time (sec)

Rat

e o

f re

acti

on


The rate will ultimately flatten The rate will ultimately flatten and the concentrations will and the concentrations will remain constant.remain constant.This is called equilibriumThis is called equilibriumΔΔG=0G=0Le Chatelier's principal etc.Le Chatelier's principal etc.More on that laterMore on that later

Rate equationRate equationThe rate equation relates the rate to the The rate equation relates the rate to the concentration of reactants.concentration of reactants.For the above example it would be For the above example it would be expressed:expressed:

Rate = k[NORate = k[NO22]]nn

The value of “k” is a constant unique to The value of “k” is a constant unique to every experiement.every experiement.The value of “n” is 0, a whole #, or a The value of “n” is 0, a whole #, or a fraction.fraction.BOTH “n” and “k” MUST BE BOTH “n” and “k” MUST BE DETERMINED BY MANIPULATION OF DETERMINED BY MANIPULATION OF EXPERIMENTAL DATA!!!EXPERIMENTAL DATA!!!I’ll show you how.I’ll show you how.

Concentration vs. RateConcentration vs. RateIf you use excel this is really easy.If you use excel this is really easy.

The computer does all the work for you.The computer does all the work for you.

I’ll show you how right now.I’ll show you how right now.

Inst. Rate vs. [NO2]

y = 0.0031x - 8E-06

R2 = 0.9778

0.00E+00

5.00E-06

1.00E-05

1.50E-05

2.00E-05

2.50E-05

0 0.002 0.004 0.006 0.008 0.01

[NO2]

Rat

e o

f re

acti

on

Doing this by Doing this by hand is not hard, hand is not hard, but it’s very but it’s very tedious.tedious.

Learn to use Learn to use excel!excel!

Zero or first order?Zero or first order?

Rate = k[NORate = k[NO22]]nn

A zero order equation has an “n” A zero order equation has an “n” value of zero. Hence, rate = kvalue of zero. Hence, rate = k

This would give a linear This would give a linear concentration vs. rate graphconcentration vs. rate graph

A 1A 1stst order reaction has an “n” value order reaction has an “n” value of 1. Hence the concentration vs. of 1. Hence the concentration vs. rate graph is linear.rate graph is linear.

Here’s another exampleHere’s another exampleIs this first order.Is this first order.

[N2O5] vs. Time

0

0.2

0.4

0.6

0.8

1

1.2

0 500 1000 1500 2000

Time (sec)

[N2O

5]

Time (sec) [N2O5]

0 1

200 0.88

400 0.78

600 0.69

800 0.61

1000 0.54

1200 0.48

1400 0.43

1600 0.38

1800 0.34

2000 0.3

Use Excel or your calculatorUse Excel or your calculatorSadly, you won’t have access to excel on the Sadly, you won’t have access to excel on the AP exam. Excel makes it easy to calculate AP exam. Excel makes it easy to calculate slope (rate)slope (rate)

Time [N2O5]

0 1

200 0.88

400 0.78

600 0.69

800 0.61

1000 0.54

1200 0.48

1400 0.43

1600 0.38

1800 0.34

2000 0.3

[N2O5] Inst. Rate

0.94 -6.00E-04

0.83 -5.00E-04

0.735 -4.50E-04

0.65 -4.00E-04

0.575 -3.50E-04

0.51 -3.00E-04

0.455 -2.50E-04

0.405 -2.50E-04

0.36 -2.00E-04

0.32 -2.00E-04Note that on the concentration vs. rate table the relationship is Note that on the concentration vs. rate table the relationship is directly proportional. 0.83 directly proportional. 0.83 -5.00E-4 and 0.405 -5.00E-4 and 0.405 -2.50E-4 -2.50E-4This show that when the concentration is halved so to is the rate.This show that when the concentration is halved so to is the rate.

If you don’t have excelIf you don’t have excelYou need to calculate the slope using your You need to calculate the slope using your calculator.calculator.This is tedious, but you need to do it.This is tedious, but you need to do it.

Rate = k[NRate = k[N22OO55]]nn

The directly proportional relationship we The directly proportional relationship we just established means that “n” = 1.just established means that “n” = 1.““k” can be established as well, it’s the k” can be established as well, it’s the slope of the concentration vs. rate graph.slope of the concentration vs. rate graph.You NEED to get good at calculating slope.You NEED to get good at calculating slope.There is another way to determine “n”…There is another way to determine “n”…

The Method of initial rateThe Method of initial rateReactions can be set up so that the Reactions can be set up so that the reverse reaction will be initially negligible.reverse reaction will be initially negligible.

NHNH44++

(aq)(aq) + NO + NO22--(aq)(aq) N N2(g)2(g) + 2H + 2H22OO(l)(l)

Note, the NNote, the N2(g)2(g) will escape from the will escape from the reaction vessel into the atmosphere.reaction vessel into the atmosphere.

In accordance with Le Chatelier's Principal In accordance with Le Chatelier's Principal the reaction will continue to be driven to the reaction will continue to be driven to the right until all the reactants are spent.the right until all the reactants are spent.

Thus the reaction rate should be Thus the reaction rate should be unaffected by the reverse reaction early in unaffected by the reverse reaction early in the reaction.the reaction.

What can you control???What can you control???You can control the initial concentrations of You can control the initial concentrations of all reactants at the start of the reaction, all reactants at the start of the reaction, and determine the rate of the creation.and determine the rate of the creation.

Consider the data below:Consider the data below:

Experiment [NH4+]0 [NO2

-]0 Initial Rate

1 0.100 0.005 1.35E-07

2 0.100 0.010 2.70E-07

3 0.200 0.010 5.40E-07

Note: A [concentration]0 means initial concentration

How do I interpret this?How do I interpret this?

Compare experiments where only the Compare experiments where only the concentration of ONLY ONE reactant is changed.concentration of ONLY ONE reactant is changed.

You can only compare Exp1 to Exp 2 and Exp 2 to You can only compare Exp1 to Exp 2 and Exp 2 to Exp 3.Exp 3.

Exp1 CANNOT be compared to Exp 3Exp1 CANNOT be compared to Exp 3


-]0 Initial Rate

1 0.100 0.005 1.35E-07

2 0.100 0.010 2.70E-07

3 0.200 0.010 5.40E-07


The rate equation will look like this:The rate equation will look like this:

Rate = k[NHRate = k[NH44++]]nn[NO[NO22

--]]mm

Note: When Exp1 is compared to Exp2 you find that Note: When Exp1 is compared to Exp2 you find that when [NOwhen [NO22

--] is doubled so to is rate. Hence [NO] is doubled so to is rate. Hence [NO22--] ]

and rate are directly proportional and m=1.and rate are directly proportional and m=1.

Because both [NHBecause both [NH44++] and [NO] and [NO22

--] change from Exp1 ] change from Exp1 to Exp3 the two can’t be compared.to Exp3 the two can’t be compared.

Use Exp2 and Exp3 to determine nUse Exp2 and Exp3 to determine n


-]0 Initial Rate

1 0.100 0.005 1.35E-07

2 0.100 0.010 2.70E-07

3 0.200 0.010 5.40E-07


1.1. What happens to [NHWhat happens to [NH44++]?]?

2.2. What happens to rate?What happens to rate?

3.3. What does that tell us about n?What does that tell us about n?

4.4. What is the final form of the rate equation?What is the final form of the rate equation?


-]0 Initial Rate

1 0.100 0.005 1.35E-07

2 0.100 0.010 2.70E-07

3 0.200 0.010 5.40E-07

The total order of the equationThe total order of the equation

Finding the total order of the rate Finding the total order of the rate equation is easy.equation is easy.

It’s simply the sum of all the It’s simply the sum of all the exponents.exponents.

Since n=1, and m=1, the total order Since n=1, and m=1, the total order of the equation is 2.of the equation is 2.

It’s a second order equation.It’s a second order equation.

You practiceYou practiceUse the table below to determine n,m,p and Use the table below to determine n,m,p and the overall order of the reaction below:the overall order of the reaction below:

BrOBrO33--(aq)(aq)+ 5Br+ 5Br--

(aq)(aq)+ 6H+ 6H++(aq)(aq)3Br3Br2(l)2(l)+3H+3H22OO(l)(l)

Experiment [BrO3-]0 [Br-]0 [H+]0 Initial Rate

1 0.10 0.10 0.10 8.00E-04

2 0.20 0.10 0.10 1.60E-03

3 0.20 0.20 0.10 3.20E-03

4 0.10 0.10 0.20 3.20E-03

Rate= k[BrORate= k[BrO33--]]nn[Br[Br--]]mm[H[H++]]pp

Integrated rate equationIntegrated rate equationWe can calculate rate, and hence we We can calculate rate, and hence we can produce a rate vs. time graph.can produce a rate vs. time graph.When it comes to rate vs. time When it comes to rate vs. time graphs, it’s important to find a graphs, it’s important to find a relationship between rate and time.relationship between rate and time.This can be accomplished in several This can be accomplished in several ways.ways.The easiest way is to use excel.The easiest way is to use excel.I’ll show you.I’ll show you.

HarderHarder

You should also memorize how You should also memorize how different graphs can be different graphs can be “straightened”.“straightened”.

I’ll show some examples on slides to I’ll show some examples on slides to come.come.

These relationships must be These relationships must be memorized, check the hall…memorized, check the hall…

Linear (zero order)Linear (zero order)

The linear graph is easy to spot, the lines are The linear graph is easy to spot, the lines are straight.straight.

This is a zero order relationshipThis is a zero order relationship


0

0.2

0.4

0.6

0.8

1

1.2

0 2 4 6 8 10

Time (sec)

Rate (grow)

Rate (decay)

Exponential (1Exponential (1stst order) order)This exponential graph is harder to spot.This exponential graph is harder to spot.

It looks a lot like the inverse graph on the It looks a lot like the inverse graph on the next slide.next slide.


1

21

41

61

0 2 4 6 8 10

Time (sec)

Co

nce

ntr

atio

n

Inverse (2Inverse (2ndnd order) order)

As you can see, this look similar to the As you can see, this look similar to the previous graph.previous graph.

When confronted with a rate vs. time graph When confronted with a rate vs. time graph like this, you need to “straighten” it. We’ll do like this, you need to “straighten” it. We’ll do this in the library tomorrow.this in the library tomorrow.

Concentration vs. time

0

5

10

15

0 2 4 6 8 10

Time (sec)

Co

nce

ntr

atio

n

Decay

Grow

Straighten your lineStraighten your line

This takes time, but you need know This takes time, but you need know how to do it.how to do it.

It’s easy with excel, but it’s a pain to It’s easy with excel, but it’s a pain to do it with your calculator.do it with your calculator.

Just be glad you don’t need to use a Just be glad you don’t need to use a slide ruler.slide ruler.

Not that I know what that is, I’m only Not that I know what that is, I’m only 27.27.

Consider the data from beforeConsider the data from beforeYou can calculate You can calculate the rate and make the rate and make a rate vs. time a rate vs. time graph.graph.

If it’s linear, then If it’s linear, then you have a first you have a first order equation.order equation.

You can also You can also calculate ln[Ncalculate ln[N22OO55] ] and graph that.and graph that.

Time (sec) [N2O5]

0 1

200 0.88

400 0.78

600 0.69

800 0.61

1000 0.54

1200 0.48

1400 0.43

1600 0.38

1800 0.34

2000 0.3

Manipulated concentration vs. timeManipulated concentration vs. time

If a graph of ln[con.] If a graph of ln[con.] vs. time is linear then vs. time is linear then the reaction has 1the reaction has 1stst order kinetics.order kinetics.

Why?Why?

CalculusCalculusFor first order equationsFor first order equations

ln[con.]=-kt+ln[con.]ln[con.]=-kt+ln[con.]00

ln is the natural logln is the natural log

-k is a constant-k is a constant

t is timet is time

Time [N2O5] Ln[N2O5] 1/[N2O5]

0 1 0 1

200 0.88 -0.12783 1.136364

400 0.78 -0.24846 1.282051

600 0.69 -0.37106 1.449275

800 0.61 -0.4943 1.639344

1000 0.54 -0.61619 1.851852

1200 0.48 -0.73397 2.083333

1400 0.43 -0.84397 2.325581

1600 0.38 -0.96758 2.631579

1800 0.34 -1.07881 2.941176

2000 0.3 -1.20397 3.333333

When it’s graphedWhen it’s graphedNote that this Note that this graph is a straight.graph is a straight.

You can prove this You can prove this by calculating the by calculating the slope of the first slope of the first two and final two two and final two points. If they’re points. If they’re the same you’re the same you’re good.good.

If that doesn’t work If that doesn’t work plot time vs. plot time vs. 1/concentration.*1/concentration.*

Ln[N2O5] vs. time

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0 500 1000 1500 2000 2500

Time (sec)

Ln

[N2O

5] (

un

itle

ss)** If that doesn’t work If that doesn’t work

you need to check your you need to check your work because I’ll only work because I’ll only give you 0, 1give you 0, 1stst, or 2, or 2ndnd order reactions.order reactions.

Look at the 1Look at the 1stst order equation order equation

ln[con.]=-kt+ln[con.]ln[con.]=-kt+ln[con.]00

Memorize this!!!Memorize this!!!

Note that this is of the y=mx+b Note that this is of the y=mx+b formatformat

-k is the slope and ln[con.]-k is the slope and ln[con.]00 is b is b

This is useful because once you This is useful because once you establish the order of the equation you establish the order of the equation you can determine the concentration at can determine the concentration at any time.any time.

And now… second orderAnd now… second order

Memorize this!!!Memorize this!!!

Note that this is of the y=mx+b formatNote that this is of the y=mx+b format

k is the slope and 1/[con.]k is the slope and 1/[con.]00 is b is b

This is useful because once you establish This is useful because once you establish the order of the equation you can the order of the equation you can determine the concentration at any time.determine the concentration at any time.

0[con]

1kt

[con]

1

First or second order?First or second order?Is this a first or Is this a first or second order second order reaction?reaction?

1.1. Create a ln[CCreate a ln[C44HH66] ] and a 1/[Cand a 1/[C44HH66]]

2.2. Graph both vs. Graph both vs. time.time.

3.3. Choose the linear Choose the linear one.one.

Time (sec) [C4H6]

0 0.01000

1000 0.00625

1800 0.00476

2800 0.00370

3600 0.00313

4400 0.00270

5200 0.00241

6200 0.00208

What is the rate constant?What is the rate constant?Now that you’ve established that it’s a Now that you’ve established that it’s a second order equation:second order equation:

1.1. Plug in 0.01000 for [CPlug in 0.01000 for [C44HH66]]00

2.2. Plug in couple of values from the table Plug in couple of values from the table for t and [Cfor t and [C44HH66]]

3.3. Solve for k using math, it’s not that hardSolve for k using math, it’s not that hard

4.4. Or find slope of the line, that’s k as wellOr find slope of the line, that’s k as well

06464 ]H[C

1kt

]H[C

1

Half lifeHalf life

Half life is defined as the amount of Half life is defined as the amount of time it takes for half of a sample to time it takes for half of a sample to decompose/decay.decompose/decay.

There are equations on page 578 There are equations on page 578 that will allow you to calculate half that will allow you to calculate half life if you have k and [con]life if you have k and [con]00

mr. bartelt presents how fast??? chapter 12 chemical kinetics

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